The two trains of lengths 400 m, 600 m respectively, running at same directions. The faster train can cross the slower train in 180 sec, the speed of the slower train is 48 km. then find the speed of the faster train?
Answer: B
Length of the two trains = 600m + 400m
Speed of the first train = X
Speed of the second train= 48 Kmph
1000/X - 48 = 180
1000/x - 48 * 5/18 = 180
50 = 9X - 120
X = 68 Kmph
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Stations A and B are 220 km apart. A train leaves towards station B with speed of 40 km/hr at 10 pm. Another train leaves towards station A from station B with speed of 50 km/hr at 11 pm. At what time both trains meet?
Answer: C
The first train has a one-hour head start.
\[ \text{covered by 11 pm} = 40 \times 1 = 40 \text{ km} \]
\[ \text{gap remaining} = 220 - 40 = 180 \text{ km} \]
From 11 pm they approach at \( 40 + 50 = 90 \) km/hr:
\[ t = \frac{180}{90} = 2 \text{ hours} \]
So they meet at 1 am. Ignoring the head start would give a meeting at about 12:27 am.
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A train speeds past a pole in 15 sec and a platform 100 m long in 25 sec, its length is?
Answer: B
The pole crossing covers the train alone, the platform crossing the train plus 100 m.
\[ L = 15v, \qquad L + 100 = 25v \]
Subtracting:
\[ 100 = 10v \quad\Longrightarrow\quad v = 10 \text{ m/s} \]
\[ L = 15 \times 10 = 150 \text{ m} \]
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A train runs across a post in 15 seconds and across a platform of length 100m in 25 seconds. Determine the length of this train:
Answer: D
Let us assume for this question that the length of train is x metres and it is assumed to be running at the speed of y m/sec.
A pole is assumed as a point object.
=>Time taken by the train to pass the pole= x/y
=>15=x/y
=>y=x/15
Now, the train passes the platform which is 100 m in length in 25 seconds.
=>x+100/y=25
=>x+100/25=y
Equating speed generated from both cases:
x+100/25=x/15
Therefore x = 150 metres.
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A jogger running at 9 kmph alongside a railway track in 240 metres ahead of the engine of a 120 metres long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?
Answer: C
The jogger runs the same way, so the train closes at the difference of the speeds.
\[ \text{relative speed} = 45 - 9 = 36 \text{ kmph} = 36 \times \frac{5}{18} = 10 \text{ m/s} \]
To pass him completely, the train must cover the 240 m gap plus its own 120 m:
\[ t = \frac{240 + 120}{10} = 36 \text{ seconds} \]
Covering only the 240 m gap brings the engine level with the jogger, not the whole train past him.
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Two trains each of which is 100 m long moving in opposite direction to one another cross each other taking 8 seconds. If speed of one train is twice the speed of other
train find the speed of the faster train.
Answer: C
Trains moving in the same direction close at the difference of their speeds; in opposite directions, at the sum.
\[ \text{relative speed} = \frac{100 + 100}{8} = 25 \text{ m/s} = 25 \times \frac{18}{5} = 90 \text{ km/hr} \]
With speeds \( v \) and \( 2v \):
\[ 3v = 90 \quad\Longrightarrow\quad v = 30 \text{ km/hr} \]
\[ \text{faster train} = 60 \text{ km/hr} \]
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The length of the bridge, which a train 130 m long and traveling at 45 km/hr can cross in 30 sec is?
Answer: C
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 45 \text{ km/hr} = 12.5 \text{ m/s} \]
\[ \text{total distance} = 12.5 \times 30 = 375 \text{ m} \]
\[ \text{bridge} = 375 - 130 = 245 \text{ m} \]
That total covers the train plus the bridge, so the train's own length is subtracted at the end.
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A train passes a station platform in 36 seconds and a man standing on the platform in 20 seconds. If the speed of the train is 54 km/hr, what is the length of the platform?
Answer: C
Speed = [54 * 5/18] m/sec = 15 m/sec.
Length of the train = (15 * 20) m = 300 m.
Let the length of the platform be x meters.
Then, x + 300 / 36 = 15
x + 300 = 540
x = 240 m.
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A goods train runs at the speed of 72 kmph and crosses a 250 m long platform in 26 seconds. What is the length of the goods train?
Answer: D
\[ 1 \text{ km/hr} = \frac{5}{18} \text{ m/s} \]
\[ 72 \text{ kmph} = 20 \text{ m/s} \]
\[ \text{total distance} = 20 \times 26 = 520 \text{ m} \]
\[ \text{train} = 520 - 250 = 270 \text{ m} \]
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There are 2 trains P and Q moving in same direction. They are of equal length and cross a stationary pole in 5 seconds and 6 seconds respectively. In how much time would they cross each other?
Answer: C
Let each train be \( L \) metres long. Crossing a pole gives each speed:
\[ v_{P} = \frac{L}{5}, \qquad v_{Q} = \frac{L}{6} \]
Moving the same way, they close at the difference:
\[ v_{P} - v_{Q} = \frac{L}{5} - \frac{L}{6} = \frac{L}{30} \]
To cross, the faster must cover both lengths, \( 2L \):
\[ t = \frac{2L}{L/30} = 60 \text{ seconds} \]
The length cancels, so it never needs a value.
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